Question

A water sample has a pH of 8.35 and a total calcium concentration of 1.63 ppm. What is the concentration of HCO3- in moles per liter? [do not neglect [H+], [OH-]]

Answer 1

a)

HCO3- + H2O <-> H+ + CO3-2

assume

pH = pKa + log(CO3-2/HCO3-)

pKa2 = 10.33

if pH = 8.35, then

8.35 = 10.33 + log(CO3-2/HCO3-)

(CO3-2/HCO3-) = 10^(8.35-10.33) = 0.01047

then

CaCO3 = Ca+2 + CO3-2

CO3-2+ H2O -- >HCO3+ + H+

assume initially,

[Ca+2] = 1.63 ppm = 1.63 mg / L = (1.63*10^-3/40) / L = 0.00004075 mol of Ca+2 / L

[CO3-2] = 0.00004075initially

now..

[CO3-2] + [HCO3-] = 0.00004075 M

[CO3-2]/[HCO3-] = 0.01047

0.01047 *[HCO3-] + [HCO3-] = 0.00004075 M

[HCO3-] = 0.00004032 M